[LeetCode] 450. Delete Node in a BST

Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST.

Basically, the deletion can be divided into two stages:

1. Search for a node to remove.
2. If the node is found, delete the node.

Note: Time complexity should be O(height of tree).

Example:

root = [5,3,6,2,4,null,7]
key = 3

    5
   / \
  3   6
 / \   \
2   4   7

Given key to delete is 3. So we find the node with value 3 and delete it.

One valid answer is [5,4,6,2,null,null,7], shown in the following BST.

    5
   / \
  4   6
 /     \
2       7

Another valid answer is [5,2,6,null,4,null,7].

    5
   / \
  2   6
   \   \
    4   7

Thought process:
Recursively search for the node to be deleted. There are several cases:

• Node.left == null && node.right == null: return null.
• Node.left == null: return node.right.
• Node.right == null: return node.left.
• Otherwise, get the minimum of the node's right sub-tree. Set node.val = minimum. And recursively delete the minimum node in the right sub-tree.

Solution:

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/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public TreeNode deleteNode(TreeNode root, int key) { if (root == null) { return null; } if (key < root.val) { root.left = deleteNode(root.left, key); } else if (key > root.val) { root.right = deleteNode(root.right, key); } else { if (root.left == null) { return root.right; } if (root.right == null) { return root.left; } TreeNode min = getMinimum(root.right); root.val = min.val; root.right = deleteNode(root.right, root.val); } return root; } private TreeNode getMinimum(TreeNode node) { while (node.left != null) { node = node.left; } return node; } }

Time complexity: O(logn).